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For the function y = f(x) = 1x³ +4: df dx ƒ'(−7) = 147 a. Find at x = -7. b. Find a formula for x = f¯¹(y). f (3) c. Find -1 df-¹ dy at y = f(-7). (ƒ¯¹)'(ƒ(−7)) = } - ( - ½ · (–339 — 4) or 23

For the function y = f(x) = 1x³ +4: df dx ƒ'(−7) = 147 a. Find at x = -7. b. Find a formula for x = f¯¹(y). f (3) c. Find -1 df-¹ dy at y = f(-7). (ƒ¯¹)'(ƒ(−7)) = } - ( - ½ · (–339 — 4) or 23

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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For the function y = f(x) = 1x³ +4: df a. Find at x = dx f'(-7) = 147 b. Find a formula for x = = f-(g). f 1(y)= c. Find df-1 dy -7. at y = f(-7). (ƒ¯¹)'(ƒ(−7)) = 1 3 · (–339 — 4) 8 -3 > ४ 8
Transcribed Image Text:For the function y = f(x) = 1x³ +4: df a. Find at x = dx f'(-7) = 147 b. Find a formula for x = = f-(g). f 1(y)= c. Find df-1 dy -7. at y = f(-7). (ƒ¯¹)'(ƒ(−7)) = 1 3 · (–339 — 4) 8 -3 > ४ 8
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